Let the vectors a⃗=(1+t) î+(1-t) ĵ+k̂, b⃗=(1-t) î+(1+t) ĵ+2 k̂ and c⃗=t î-t ĵ+k̂, t ∈ R be such that for α, β, γ ∈ R, α a⃗+β b⃗+γ c⃗=0⃗ ⇒ α=β=γ=0. Then, the set of all values of t is :

Solution

Maths sir · 4 steps

Only the trivial combination gives zero, so the vectors are independent: make the determinant non-zero.

1. Read the condition. α a⃗+β b⃗+γ c⃗=0⃗ forcing α=β=γ=0 means the three vectors are linearly independent, so they are not coplanar. For three vectors in space, that is the same as [a⃗ b⃗ c⃗] ≠ 0.
2. Simplify first. Do C₁ → C₁ + C₂. The first column becomes 2, 2, 0 and every t in it cancels. Expanding along the new first column is now short.

   |1+t, 1-t, 1; 1-t, 1+t, 2; t, -t, 1| = |2, 1-t, 1; 2, 1+t, 2; 0, -t, 1| = 6t

3. Expand. 2[(1+t) + 2t] - 2[(1-t) + t] = 2 + 6t - 2 = 6t. Check t = 0: a⃗ = î+ĵ+k̂, c⃗ = k̂ and b⃗ = î+ĵ+2k̂ = a⃗+c⃗, so the vectors really are dependent there.
4. So. 6t ≠ 0 exactly when t ≠ 0. The set of all values of t is R-{0}, option (c).

The idea to keep. Only the trivial combination gives 0⃗ means the vectors are linearly independent, so the box product is non-zero. Find where it is zero, then remove those values from R.

Answer from the NTA final answer key. Solution by PrepWiser.